Expectation value of two annihilation operators

In summary, the conversation discusses the effect of a beam splitter on a quantum optical system and how it affects the measurement of intensity using a photodiode. The operators for the outgoing beams are represented by c = (a + ib)/√2 and the intensity is proportional to <c^{\dag} c>. The book explains that this can be written as <a^{\dag} a> + <b^{\dag} b> + i(<a^{\dag}><b> - <b^{\dag}><a>)/2, raising a question about whether this is always true for any two commuting operators.
  • #1
Ancient_Nomad
15
0
Hello,

I was studying about the effect of a beam splitter in a text on quantum optics. I understand that if a and b represent the mode operators for the two beams incident on the splitter, then the operator for one of the outgoing beams is the following,
[tex] c = \frac{(a + ib)}{\sqrt{2}} [/tex]​

Now if I try to measure the intensity of this beam by a photodiode, the intensity will be proportional to-

[tex]<c^{\dag} c>[/tex]

On evaluating this, I get,
[tex]\frac{\left(<a^{\dag} a> + <b^{\dag} b> + i(<a^{\dag} b> - <b^{\dag} a>)\right)}{2} [/tex]
Now the book says, that this can be written as,
[tex]\frac{\left(<a^{\dag} a> + <b^{\dag} b> + i(<a^{\dag}><b> - <b^{\dag}><a>)\right)}{2} [/tex]

I am unable to understand this step, that is [tex]<a^{\dag} b> = <a^{\dag}><b>[/tex]
can someone please explain this.
I understand that these mode operators commute, but is this always true for any two commuting operators.

Thanks
 
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  • #2
if you write what [itex] < a^{\dagger} \, b > [/itex] is, then I am sure that you can figure it out.
 
  • #3
for your question. I can provide an explanation for the expectation value of two annihilation operators.

First, let's define what an annihilation operator is. An annihilation operator is a mathematical operator that is used in quantum mechanics to describe the destruction or annihilation of a particle. In this case, the a and b operators represent the modes of the two beams incident on the beam splitter.

Now, the c operator that you have mentioned in your question is a combination of the a and b operators, which represents the output of one of the beams after passing through the beam splitter. The intensity of this beam can be measured by the photodiode, and the expected value of this measurement is given by <c^{\dag} c>.

In order to calculate this expected value, we need to use the properties of annihilation operators. One of these properties is that they commute with their corresponding creation operators, which in this case are a^{\dag} and b^{\dag}. This means that <a^{\dag} a> = <a><a^{\dag}> = <a^{\dag}><a>, and the same for b^{\dag}.

Using this property, we can rewrite the expression for the expected value as follows:

<c^{\dag} c> = <\frac{(a + ib)^{\dag}}{\sqrt{2}} \frac{(a + ib)}{\sqrt{2}}> = <\frac{(a^{\dag} - ib^{\dag}) (a + ib)}{2}> = <\frac{a^{\dag} a + b^{\dag} b + i(a^{\dag} b - b^{\dag} a)}{2}>

Now, using the property that annihilation and creation operators commute, we can write <a^{\dag} b> = <a^{\dag}><b> and <b^{\dag} a> = <b^{\dag}><a>. This results in the expression given in the book:

<c^{\dag} c> = <\frac{a^{\dag} a + b^{\dag} b + i(<a^{\dag}><b> - <b^{\dag}><a>)}{2}>

In summary, the expectation value of two annihilation operators can be calculated by using the properties of annihilation and creation operators, which include commutativity. This is true for any
 

What is the expectation value of two annihilation operators?

The expectation value of two annihilation operators is a mathematical quantity that represents the average value of a quantum observable in a given state. It is calculated by taking the inner product of the state vector and the operator, and then normalizing it by the state's norm.

Why is the expectation value of two annihilation operators important in quantum mechanics?

The expectation value of two annihilation operators is important because it allows us to predict the outcome of measurements in quantum systems. It also plays a crucial role in the formulation of quantum field theory, which is used to describe the behavior of particles at a subatomic level.

How is the expectation value of two annihilation operators related to the uncertainty principle?

The expectation value of two annihilation operators is related to the uncertainty principle through the commutation relation between the operators. This relation states that the product of the uncertainties in two non-commuting observables, such as position and momentum, cannot be smaller than a certain value. This relationship is fundamental to the unpredictability of quantum systems.

What is the difference between the expectation value of two annihilation operators and the expectation value of one annihilation operator?

The expectation value of two annihilation operators represents the average value of two observables, while the expectation value of one annihilation operator represents the average value of a single observable. In some cases, the expectation value of two annihilation operators may be equal to the expectation value of one annihilation operator, but in general, they are distinct quantities with different physical interpretations.

Can the expectation value of two annihilation operators be negative?

Yes, the expectation value of two annihilation operators can be negative. This can occur when the state vector and the operator have opposite signs, resulting in a negative inner product. In quantum mechanics, negative expectation values are common and do not necessarily indicate a physically impossible result.

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